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Let $X$ be a topological space. Let $f \in C(X)$. Define the support of $f$. $$\operatorname{supp}(f) := \overline{\{x \in X: f(x) \neq 0\}}$$

I want to show that $$A:=\{f \in C(X)\mid \exists K \subseteq X \mathrm{\ compact \ }: \forall x \notin K: f(x) = 0\}$$

and $$B:=\{f \in C(X): \operatorname{supp}(f) \mathrm{\ is \ compact}\}$$

coincide. The inclusion $B \subseteq A$ is trivial.

Attempt: I managed to show that $A \subseteq B$ if $X$ is Hausdorff:

If $f \in A$, determine a compact set $K$ with $f(x) = 0$ for $x \notin K$. Then $$\{x \in X: f(x) \neq 0\} \subseteq K$$ and taking closures $$\operatorname{supp}(f) \subseteq \overline{K}$$ Since $X$ is Hausdorff, $K$ is closed so we get $$\operatorname{supp}(f) \subseteq K$$

Then $\operatorname{supp}(f)$ is a closed subset of a compact set and we can conclude.

Question: Does the inclusion remain valid without the Hausdorff assumption?

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    $\begingroup$ This is a very good question in my opinion. $\endgroup$ – JustDroppedIn Jul 7 at 10:51
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Not necessarily. You can take as an example a modification of the line with two zeros - in particular, let $C$ be an infinite set and let $$X=C\cup(0,1].$$ Define $f:X\rightarrow\mathbb R$ by $$f(x) = \begin{cases}0 & \text{if }x\in C\\x & \text{if }x\in (0,1].\end{cases}$$ Define a topology on $X$ by saying that a set $U\subseteq X$ is open if and only if $f(U)$ is open. This really does define a topology, though you should carefully check what happens with intersections of open sets.

Clearly, $f$ is continuous. Moreover, if $c\in C$, then $\{c\}\cup (0,1]$ is homeomorphic by $f$ to $[0,1]$ - meaning that $f\in A$ since this is a compact set containing all non-zero values of $f$. However, the support of $f$ is all of $X$, which is not compact because the cover $\{\{c\} \cup (0,1] : c\in C\}$ has no non-trivial subcovers (and hence no finite subcovers). Thus $f\not\in B$.

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